MAM1000 – Page 10 – Mathemafrica

UCT MAM1000 lecture notes part 20 – part ii – Monday 17th August

So, we saw in the last section that we could work out the polynomial expression for $(1+x)^5$ both using combinatorics as well as using calculus. We had also found previously that for small $x$ we could just take the first couple of terms of the polynomial and it was a good approximation of the function itself, depending on how accurate we needed the answer. For instance, for small $x$ , $1+5x$ is a reasonable approximation. One thing to note is that the value of these two functions is exactly the same at $x=0$ and the derivative of both functions is the same at $x=0$ . The second derivatives are not the same, but had we taken the next term in the polynomial, the second derivatives would have matched as well.

OK, let’s take this one step further and repeat everything exactly as before, but now with the power, not as 5 but as 5.2 and see what happens.…

By Jonathan Shock| August 16th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|1 Comment

UCT MAM1000 lecture notes part 20 – part i – Monday 17th August

The beginnings of Taylor polynomials and approximations

We’ll start off with some of the discussion we had in class today about treating the binomial expansion as a way to approximate a function.

We know that $(x+y)^5=\sum_{r=0}^5{_5C_r} x^r y^{5-r}$ . We can then see that if we have a function like $(1+x)^5$ we can simply replace the $y$ in the above expression by 1 and we get (writing out the sum explicitly):

$(1+x)^5=1+5x+10 x^2+10x^3+5 x^4+x^5$

If you can’t remember these binomial coefficients offhand, just write out the first 6 lines of Pascal’s triangle and read them off.

ok, so what if I asked you to give me an approximate value for $(1+0.01)^5$ . This isn’t so easy because you’ll have to multiply 1.01 by itself five times. How about if we use the polynomial form from the binomial expansion? Well, we can see that this is:

$(1+0.01)^5=1+5(0.01)+10 (0.01)^2+10(0.01)^3+5 (0.01)^4+(0.01)^5$

But we can also see that each term gets smaller and smaller.…

By Jonathan Shock| August 14th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|1 Comment

UCT MAM1000 lecture notes part 19

The Binomial expansion

So, we’re going to start using some of the combinatorics we’ve just learned to answer some questions which without these techniques would seem completely infeasible.

What is $(x+y)^{34}$ ? This seems like an almost impossible task; you’d have to write out $(x+y)(x+y)(x+y)...$ , 34 times, then multiply them all out and it would become incredibly messy. However, everything we’ve done in the last section will allow us to see precisely what this, and any other expression of this form, is given by. Let’s start with a simpler example:

How about $(x+y)^4$ ? This is a simpler example and one that we could think of doing by hand, but we will show that there is a very general way to get any expression of this form. We start by writing this out in long form:

$(x+y)^4=(x+y)(x+y)(x+y)(x+y)$

We are now going to label the $x$ ‘s and $y$ ‘s, though the labels (which will be indices) will just be dummy labels and we will remove them in the end.…

By Jonathan Shock| August 10th, 2015|Courses, First year, MAM1000, Uncategorized, Undergraduate|1 Comment

UCT MAM1000 lecture notes part 18 – Friday 14th August

How to choose $r$ objects from $n$ objects

Let’s say that we have 6 objects ( ${\{a,b,c,d,e,f\}}$ ) and we want to know the number of ways of picking 4 of them (where the order doesn’t matter, so ${\{a,b,c,d\}}$ is no different from ${\{a,b,d,c\}}$ ). We can imagine writing down all of the possible permutations of the 6 objects (of which there are ${6!}$ as we know) and then just taking the first four items. For instance, these are the first few permutations and the items in brackets are the ones that we pick as our 4:

$\begin{array}{c} \text{(abcd)ef} \\ \text{(abcd)fe} \\ \text{(abce)df} \\ \text{(abce)fd} \\ \text{(abcf)de} \\ \text{(abcf)ed} \\ \text{(abdc)ef} \\ \text{(abdc)fe} \\ \text{(abde)cf} \\ \text{(abde)fc} \\ ... \\ \end{array}$

If we pick the items in the brackets then clearly we are going to get many repetitions of the same thing. For instance, the first two items are the same (both (abcd)). Also many of the items in the brackets are the same, up to a reordering, and we said that we didn’t care about the order, so (abcd) is the same as (abdc).…

By Jonathan Shock| August 9th, 2015|Courses, First year, MAM1000, Uncategorized, Undergraduate|1 Comment

UCT MAM1000 lecture notes part 17 – Thursday 13th August

Here we’re going to continue looking at how many different ways there are to order collections of objects, but with certain constraints.

We saw previously that if you have $n$ (distinguishable) objects, you can arrange them in $n!$ ways ( $n$ positions for the first object $(n-1)$ for the second, etc, with a single position for the last object).

Now we can ask what is the way of arranging $r$ of $n$ objects. That is, if I give you a deck of cards and ask you to take any five cards and put them in some order, how many possibilities are there in total?

Well, clearly the first card is going to be one of 52 different possibilities (in a 52 card deck). The second card is going to be one of 51 possibilities, etc. So the answer is going to be:

$52\times 51\times 50\times 49\times 48$

can we write this in terms of factorials? Let’s multiply and divide this number by $47!$ :

$\frac{52\times 51\times 50\times 49\times 48\times 47!}{47!}$

But the top is just $52!$ , so this is given by:

$\frac{52!}{47!}$

Where did the $47$ come from?…

By Jonathan Shock| August 9th, 2015|Courses, First year, MAM1000, Uncategorized, Undergraduate|3 Comments

UCT MAM1000 Lecture notes part 15 – Tuesday 11th August

This section is going to be a pretty diverse one, but is going to give you a very powerful set of tools for solving diverse problems in the real world.

You will be able to calculate the probability of two people in a given sized room having the same birthday, you will be able to calculate the combinatorics of all sorts of permutations and combinations, and you’ll learn about how to approximate many functions by polynomial expressions – this is an incredibly powerful technique and is used throughout the sciences.

Combinatorics

Combinatorics is the study of the different ways to rearrange objects given certain constraints. A question might be:

If a teacher is randomly picking two different students out per class for a year to get them to explain an answer on the board , what is the likelihood that you will be picked exactly five times?

How about the number of times Jodhi is likely to be called in a given game of Thunee?…

By Jonathan Shock| August 8th, 2015|Courses, First year, MAM1000, Uncategorized, Undergraduate|5 Comments

Average Value of a function

If we have a function in some range $[a,b]$ we can ask what the average value of the function is. We are going to do this in a very intuitive way. We split up the curve into $n$ points and take the average of those points, and then ask what happens to that number as we take the number of points to infinity. Let’s consider the curve $y=x^2-x^3$ between $-3$ and $3$ . If we take seven points along the curve (ie. at $x=-3,-2,-1,0,1,2,3$ ) We will get the function values: $f(x)=36, 12, 2, 0, 0, -4, -18$ . as can be seen here:

The average value of the function when we take just seven sample points is thus: $\frac{36+ 12+ 2+ 0+ 0 -4 -18}{7}=4$ . The figure below shows what happens when we take more and more points. We see that the mean value converges to a fixed number,in this case 3.

But there is a better way to do this and again, it involves integration.…

By Jonathan Shock| August 1st, 2015|Courses, First year, MAM1000, Uncategorized, Undergraduate|1 Comment

Arc lengths

We have now learnt a great deal about various properties of curves. We can study their gradients, the areas under them and between them, we can even study properties of functions which have discontinuities and which extend all the way to $\infty$ . We can look at the revolution of a curve about an axis and study the volume enclosed using cylinders of various forms and we have a good understanding of the link between finite sums of pieces which make up an area or volume and the limit that these pieces become infinitesimally small and thus how we end up with an integral from a Riemann sum.

There is one piece of the puzzle left, which is to know the length of a curve between two points. Let’s say we have some function $f(x)$ and we want to know the length of the curve (ie. how long a piece of string would be that went along the curve and stretched from some point $a$ to another point $b$ on the curve).…

By Jonathan Shock| August 1st, 2015|Courses, First year, MAM1000, Uncategorized, Undergraduate|1 Comment

Volumes by cylindrical shells

Sometimes it is easier to calculate the volume of a solid (formed by the revolution of a surface about an axis of revolution) by dividing it up in a different way than into thin cross-sections. We will still use cylinders, but this time they will be cylindrical shells, where the height of the cylindrical shell mimics the function. This figure illustrates how this works for a particular function (in this case $f(x)=\sqrt{1-x^2}$ between $x=0$ and $x=1$ .

Top Left image: the function $\sqrt{1-x^2}$ .
Top right image: The volume formed by revolving this about the $y$ axis.
Bottom left image: Breaking this shape up into cylindrical shells (think of taking a wire loop of varying diameters and slicing vertically through the function.
Bottom right image: The cylindrical shells when pulled apart (this is just for illustration purposes). In this case the central shell is taller than the outer shells because in the centre of the shape the function is highest and towards the edges it is lowest.

…

By Jonathan Shock| August 1st, 2015|Courses, First year, MAM1000, Uncategorized, Undergraduate|3 Comments

UCT MAM1000 lecture notes part 11 – Tuesday 4th August

We can choose to rotate our shapes around different axes and of course the volumes will be very different. If we choose the shape defined by the area in between the functions $y=x^2$ and $y=x$ and we choose to rotate it around the vertical line $x=-1$ , we get the shape on the left hand side of this figure:

The cross-section is shown here:

Again, we have to ask what the cylinder is going to look like at height $y$ (ask yourself what a thin slice of the shape would look like at height $y$ : It’s going to be an annulus). This time both the inner and the outer radii of the annulus are going to change as we change $y$ . Now the inner radius is going to be $1+y$ and the outer radius is going to be $1+\sqrt{y}$ .
Thus we can see that the annulus at height $y$ is going to have area $\pi ((1+\sqrt{y})^2-(1+y)^2)$ (the 1 is because we are going from the point $x=-1$ to the point $y$ and $\sqrt{y}$ ).…

By Jonathan Shock| August 1st, 2015|Courses, First year, MAM1000, Uncategorized, Undergraduate|5 Comments

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